Question

A particle moves in \(x-y\) plane according to the equations \(x=4 t^{2}+5 t+16\) and \(y=5 t\) where \(x, y\) are in metre and \(t\) is in second. The acceleration of the particle is

(a) \(8 \mathrm{~m} \mathrm{~s}^{-2}\)

(b) \(12 \mathrm{~m} \mathrm{~s}^{-2}\)

(c) \(14 \mathrm{n} \mathrm{s}^{-2}\)

(d) \(16 \mathrm{~m} \mathrm{~s}^{-2}\)

Answer 1

The correct option of this question will be (a).

Solution —

As $x=4 t^{2}+5 t+16$ and $y=5 t$

$\therefore \quad v_{x} =\frac{d x}{d t}=\frac{d}{d t}\left(4 t^{2}+5 t+16\right)=8 t+5$

$v_{y} =\frac{d y}{d t}=\frac{d}{d t}(5 t)=5$

$a_{x} =\frac{d v_{x}}{d t}=\frac{d}{d t}(8 t+5)=8$

$\text { and } a_{y} =\frac{d v_{y}}{d t}=\frac{d}{d t}(5)=0$

The acceleration of the particle is

$\vec{a}=a_{x} \hat{i}+a_{y} \hat{j}=8 \hat{i}+0 \hat{j}=8 \hat{i}$

or $a=\sqrt{8^{2}}=8 ms ^{-2}$